Q.If $X\sim B(4, p)$ and $P(X = 0) = \dfrac{16}{81}$, then $P(X = 4) = \ldots$ (A) $\dfrac{1}{16}$ (B) $\dfrac{1}{81}$ (C) $\dfrac{1}{27}$ (D) $\dfrac{1}{8}$

When a random variable $X$ counts the number of successes in $n$ independent Bernoulli trials, each with the same success probability $p$ (and failure probability $q=1-p$), $X$ is said to follow a binomial distribution with parameters $n$ and $p$, written $X\sim B(n,p)$. Its probability mass function gives the chance of exactly $x$ successes as $P(X=x)={}^{n}C_x\,p^xq^{n-x}$ for $x=0,1,2,\ldots,n$, where ${}^{n}C_x=\dfrac{n!}{x!(n-x)!}$ counts the number of ways to choose which $x$ of the $n$ trials were the successful ones, and $p^xq^{n-x}$ is the probability of any one such arrangement (by independence). These $(n+1)$ probabilities are exactly the successive terms of the binomial expansion of $(q+p)^n$, which is why they …